Base-10 Log and the Number of Digits in Numbers

Introduction

When preparing for quantitative trading interviews, you're bound to encounter seemingly simple mathematical concepts that have profound implications. One such concept is understanding why the base-10 logarithm of a number, $\log_{10}(n)$, effectively gives the number of digits in $n$ minus 1.

This idea might seem abstract at first, but it's a critical tool for mental math, time complexity analysis, and even number theory — all of which are relevant in quant trading interviews. In this post, we’ll break down why this happens, step-by-step, so you can walk into your next interview with confidence.

What Does "Number of Digits" Mean?

To understand the relationship between $\log_{10}(n)$ and the number of digits in $n$, let's first define what "number of digits" means.

The number of digits $d$ in a positive integer $n$ is the count of individual symbols (0-9) required to write $n$.

For example:

If we look at the ranges of these numbers, we can generalize that the number of digits in $n$ corresponds to the powers of 10 that "contain" $n$:

For instance:

Relating Digits to Base-10 Log

To see the link to logarithms, we can take the base-10 logarithm of the inequality:

Using the logarithmic identity $\log_{10}(10^x) = x$, this becomes:

This inequality tells us that $\log_{10}(n)$ lies between $d-1$ and $d$. In other words, the floor of $\log_{10}(n)$ (i.e., the greatest integer less than or equal to $\log_{10}(n)$) is exactly $d - 1$.

Extracting the Number of Digits from Base-10 Log

We now know that $d-1 \leq \log_{10}(n) < d$, which means that:

where $\lfloor x \rfloor$ denotes the "floor" function — the greatest integer less than or equal to $x$.

To get the number of digits $d$ from $\log_{10}(n)$, we rearrange the equation:

This equation tells us that, to compute the number of digits of $n$, we just need to take the floor of $\log_{10}(n)$ and add 1.

Example Walkthrough

Find the number of digits of $2^8$

For $n = 2^8$:

Using the approximation $\log_{10}(2) \approx 0.301$:

Thus, the number of digits is:

Indeed, $2^8 = 256$ has 3 digits.

Find the number of digits of $12^{10}$

For $n = 12^{10}$:

Using the approximation $\log_{10}(12) \approx 1.079$:

Thus, the number of digits is:

Indeed, $12^{10} = 61,917,364,224$ has 11 digits.

Find the number of digits of $137^3$

For $n = 137^3$:

Using the approximation $\log_{10}(137) \approx 2.137$:

Thus, the number of digits is:

Indeed, $137^3 = 2,571,353$ has 7 digits.

Why Does Base-10 Log "Give the Number of Digits Minus 1"?

At this point, the key observation is that since $d = \lfloor \log_{10}(n) \rfloor + 1$, it follows that:

This is why some people say that $\log_{10}(n)$ "gives the number of digits minus 1" — because the integer part of $\log_{10}(n)$ tells you "how many powers of 10 fit into $n$".

To think about it intuitively, the value of $\log_{10}(n)$ is the "power of 10" needed to reach $n$. For example:

Key Takeaways

Number of Digits

To compute the number of digits $d$ of an integer $n$, use:

This equation is essential for quant interviews where you might need to compute digit counts in your head.

Digits Minus 1

The "number of digits minus 1" is given directly by $\lfloor \log_{10}(n) \rfloor$. This tells you how many powers of 10 "fit into" $n$.

Use in Interviews

This concept is useful in mental math, especially in time complexity analysis (like the number of iterations in a binary search) and in number theory questions, where you might be asked to count digits or powers.